Решебник по алгебре 9 класс Макарычев Задание 562

\[\boxed{\text{562\ (562).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Первые\ десять\ членов\ \]

\[последовательности:\]

\[1;4;9;16;25;36;49;64;81;100.\]

\[a_{20} = 20^{2} = 400,\ \ \]

\[a_{40} = 40² = 1600,\ \ a_{n} = n².\]

\[\boxed{\text{562.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[a^{2} = x;\ \ b^{2} = x + d;\ \ c^{2} =\]

\[= x + 2d;\]

\[\frac{1}{a + c} = \frac{1}{b + c} + \left( \frac{1}{a + c} - \frac{1}{b + c} \right) =\]

\[= \frac{1}{b + c} + \frac{b - a}{(a + c)(b + c)} =\]

\[= \frac{1}{b + c} + \frac{b^{2} - a^{2}}{(a + c)(b + c)(a + b)} =\]

\[= \left( \frac{1}{b + c} \right) + \frac{d}{(a + c)(b + c)(a + b)};\]

\[\frac{1}{a + b} =\]

\[= \frac{1}{b + c} + \left( \frac{1}{a + b} - \frac{1}{b + c} \right) =\]

\[= \frac{1}{b + c} + \frac{b + c - a - b}{(a + b)(b + c)} =\]

\[= \frac{1}{b + c} + \frac{c - a}{(a + b)(b + c)} =\]

\[= \frac{1}{b + c} + \frac{c^{2} - a^{2}}{(a + c)(b + c)(a + b)} =\]

\[= \frac{1}{b + c} + \frac{2d}{(a + c)(b + c)(a + b)}.\]

\[Что\ и\ требовалось\ доказать.\]